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explain solution rd sharma class 12 chapter inverse trigonometric functions exercise 3.2 question 4 sub question (i) maths

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Answer: \frac{5\pi }{6}

Hint: Any x x\in \left [ -1,1 \right ],\cos ^{-1}represents an angle in \left ( 0,\pi \right )

Given:    \cos ^{-1}\left ( \frac{-\sqrt{3}}{2} \right )

Solution:

\cos ^{-1}\left ( \frac{-\sqrt{3}}{2} \right ) is an angle in \left [ 0,\pi \right ]whose cosine is \left ( \frac{-\sqrt{3}}{2} \right )

                                =\pi -\frac{\pi }{6}

                                =\frac{6\pi-\pi }{6}

  \cos ^{-1}\left ( \frac{-\sqrt{3}}{2} \right ) = \frac{5\pi }{6}

         

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